(a) find all points where the function f(z) = (x^2+y^2-2y)+i(2x-2xy) is differentiable, and compute the derivative at those points.

The given function is differentiable at y = 1.

At y = 1, f'(z)  = 0

Step-by-step explanation:

As per the given question,

Let z = x + i y

Suppose,

On computing the partial derivatives of u and v as:

And

According to the Cauchy-Riemann equations

Now,

Therefore,

holds only.

This means,

2y - 2 = 0

⇒ y = 1

Therefore f(z) has a chance of being differentiable only at y =1.

Now we can compute the derivative

At y = 1

f'(z) = 0

Hence, the required derivative at y = 1 ,  f'(z)  = 0

t = 3

step-by-step explanation:

put t = 3 in this expression :

(3²-1)(3²-2)(3² - 3)(3² - 4) = 8×7×6×5 = 56×30 = 1680

step-by-step explanation: